Redox reactions task 30. How to solve problems C1 (30) on the Unified State Exam in chemistry

Line UMK Kuznetsova. Chemistry (10-11) (U)

Line UMK Kuznetsova. Chemistry (10-11) (B)

Line UMK N. E. Kuznetsova. Chemistry (10-11) (basic)

Organization of preparation for the Unified State Exam in chemistry: redox reactions

The material was prepared based on the webinar “Organization of preparation for the Unified State Exam in Chemistry: redox reactions”

“We are looking at the organization of preparation for the successful completion of tasks related to redox reactions. If we look at the specification and demo version, such reactions are directly related to tasks No. 10 and No. 30, but this is a key topic in a school chemistry course. It touches on a variety of issues, a variety of properties chemical substances. It is very extensive,” emphasizes Lidia Asanova, presenter of the webinar, candidate of pedagogical sciences, author of teaching aids.

Task No. 30, considering redox reactions - task high level difficulties. To receive the highest score (3) for its completion, the student’s answer must include:

• determination of the oxidation state of elements that are oxidizing and reducing agents;
• oxidizing agent and reducing agent (elements or substances);
• processes of oxidation and reduction, and on their basis a compiled electronic (electron-ion) balance;
• determination of substances missing in the reaction equation.

However, students often skip, do not assign coefficients, do not indicate the oxidizing agent and reducing agent, and oxidation states. How should you organize work in class in order to achieve good results in the exam?

Particular attention in O. S. Gabrielyan’s textbook for grade 10, intended for studying the subject for 3–4 hours a week, is given to applied topics: the manual covers chemistry-related issues of ecology, medicine, biology and culture. In grade 11, the course is completed and summarized.

1. Preparation for the exam should be carried out in the process of teaching the subject and preparation cannot be reduced only to training in performing tasks similar to the tasks exam paper. Such “coaching” does not develop thinking or deepen understanding. But, by the way, the exam task states that other wording of the answer is allowed without distorting its meaning. This means that by creatively and understandingly approaching the solution to the task at hand, you can get the highest score for completion, even if the answer is formulated differently.

The main task of preparing for the exam is targeted work on repetition, systematization and generalization of the studied material, on bringing the key concepts of the chemistry course into the knowledge system. Of course, experience in conducting a real chemical experiment is required.

2. There is a list of topics and concepts that schoolchildren should not forget at all. Among them:

• rules for determining the oxidation states of atoms (in simple substances, the oxidation state of elements is zero, the highest (maximum) oxidation state of elements of groups II-VII, as a rule, is equal to the number of the group in which the element is located in the periodic table, the lowest (minimum) oxidation state of metals equal to zero, etc.);
• the most important oxidizing and reducing agents, and also the fact that the oxidation process is always accompanied by a reduction process;
• redox duality;
• types of ORR (intermolecular, intramolecular, comporportionation reactions, disproportionation reactions (self-oxidation-self-reduction)).

The table lists the types of redox reactions and factors influencing the course of reactions (photo pages). Examples are analyzed in detail, and, in addition, there are tasks on the topic “OVR” in the Unified State Exam format.

For example:

“Using the electron balance method, create an equation for the chemical reaction:

N 2 O + KMnO 4 + … = NO 2 + … + K 2 SO 4 + H 2 O

Specify the oxidizing agent and the reducing agent."

However, a variety of examples are given to practice solving problems. For example, in the textbook “Chemistry. Advanced level. Grade 11. Tests" are as follows:

“Based on the theory of redox processes, indicate the schemes not possible reactions.

SO 2 + H 2 S → S + H 2 O

S + H 2 SO 4 → SO 2 + H 2 O

S + H 2 SO 4 → H 2 S + H 2 O

K 2 SO 3 + K 2 Cr 2 O 7 + H 2 SO 4 → K 2 SO 4 + K 2 CrO 4 + H 2 O

KMnO 4 + HCl → Cl2 + MnCl 2 + KCl + H 2 O

I 2 + SO 2 + H 2 O → HIO 3 + H 2 SO 4

Justify your answer. Convert diagrams of possible processes into reaction equations. Specify the oxidizing agent and reducing agent"

“Make up reaction equations in accordance with the scheme of changes in the oxidation states of carbon atoms: C 0 → C – 4 → C –4 → C +4 → C +2 → C –2.”

“Given substances: carbon, nitrogen oxide (IV), sulfur oxide (IV), an aqueous solution of potassium hydroxide. Write equations for four possible reactions between these substances, without repeating pairs of reactants.”

All this allows you to study the topic of redox reactions as fully as possible and work out solutions to a variety of problems.

, awarded the Presidential Prize in the field of education. These are textbooks and manuals in subject areas that are necessary for the development of the scientific, technical and production potential of Russia.

We continue to discuss the solution to problem type C1 (No. 30), which will definitely be encountered by everyone who will take the Unified State Exam in chemistry. In the first part of the article we outlined the general algorithm for solving problem 30, in the second part we analyzed several rather complex examples.

We begin the third part with a discussion of typical oxidizing and reducing agents and their transformations in various media. Fifth step

: we discuss typical OVRs that may occur in task No. 30 I would like to recall a few points related to the concept of oxidation state. We have already noted that a constant degree of oxidation is characteristic only of a relatively small number of elements (fluorine, oxygen, alkaline and etc.) Most elements can exhibit different states of oxidation. For example, for chlorine all states are possible from -1 to +7, although odd values ​​are most stable. Nitrogen exhibits oxidation states from -3 to +5, etc.

There are two important rules to remember clearly.

1. The highest oxidation state of a non-metal element in most cases coincides with the number of the group in which the element is located, and the lowest oxidation state = group number - 8.

For example, chlorine is in group VII, therefore, its highest oxidation state = +7, and its lowest - 7 - 8 = -1. Selenium is in group VI. The highest oxidation state = +6, the lowest - (-2). Silicon is located in group IV; the corresponding values ​​are +4 and -4.

Remember that there are exceptions to this rule: the highest oxidation state of oxygen = +2 (and even this only appears in oxygen fluoride), and the highest oxidation state of fluorine = 0 (in a simple substance)!

2. Metals are not capable of exhibiting negative oxidation states. This is quite important, considering that more than 70% of chemical elements are metals.

And now the question: “Can Mn(+7) act as a reducing agent in chemical reactions?” Take your time, try to answer yourself.

Correct answer: "No, it can't!" It's very easy to explain. Take a look at the position of this element on the periodic table. Mn is in group VII, therefore its HIGH oxidation state is +7. If Mn(+7) acted as a reducing agent, its oxidation state would increase (remember the definition of a reducing agent!), but this is impossible, since it already has a maximum value. Conclusion: Mn(+7) can only be an oxidizing agent.

For the same reason, ONLY OXIDATING properties can be exhibited by S(+6), N(+5), Cr(+6), V(+5), Pb(+4), etc. Take a look at the position of these elements in periodic table and see for yourself.

And another question: “Can Se(-2) act as an oxidizing agent in chemical reactions?”

And again the answer is negative. You probably already guessed what's going on here. Selenium is in group VI, its LOWEST oxidation state is -2. Se(-2) cannot GET electrons, i.e., cannot be an oxidizing agent. If Se(-2) participates in ORR, then only in the role of a REDUCER.

For a similar reason, the ONLY REDUCING AGENT can be N(-3), P(-3), S(-2), Te(-2), I(-1), Br(-1), etc.

The final conclusion: an element in the lowest oxidation state can act in the ORR only as a reducing agent, and an element with the highest oxidation state can only act as an oxidizing agent.

"What if the element has an intermediate oxidation state?" - you ask. Well, then both its oxidation and its reduction are possible. For example, sulfur is oxidized in a reaction with oxygen, and reduced in a reaction with sodium.

It is probably logical to assume that each element in the highest oxidation state will be a pronounced oxidizing agent, and in the lowest - a strong reducing agent. In most cases this is true. For example, all compounds Mn(+7), Cr(+6), N(+5) can be classified as strong oxidizing agents. But, for example, P(+5) and C(+4) are restored with difficulty. And it is almost impossible to force Ca(+2) or Na(+1) to act as an oxidizing agent, although, formally speaking, +2 and +1 are also higher degrees oxidation.

On the contrary, many chlorine compounds (+1) are powerful oxidizing agents, although the +1 oxidation state in this case is far from the highest.

F(-1) and Cl(-1) are bad reducing agents, while their analogues (Br(-1) and I(-1)) are good. Oxygen in the lowest oxidation state (-2) exhibits practically no reducing properties, and Te(-2) is a powerful reducing agent.

We see that everything is not as obvious as we would like. In some cases, the ability to oxidize and reduce can be easily foreseen; in other cases, you just need to remember that substance X is, say, a good oxidizing agent.

It seems that we have finally reached the list of typical oxidizing and reducing agents. I would like you to not only “memorize” these formulas (although that would be nice!), but also be able to explain why this or that substance is included in the corresponding list.

Typical oxidizing agents

1. Simple substances - non-metals: F 2, O 2, O 3, Cl 2, Br 2.
2. Concentrated sulfuric acid (H 2 SO 4), nitric acid (HNO 3) in any concentration, hypochlorous acid (HClO), perchloric acid (HClO 4).
3. Potassium permanganate and potassium manganate (KMnO 4 and K 2 MnO 4), chromates and dichromates (K 2 CrO 4 and K 2 Cr 2 O 7), bismuthates (e.g. NaBiO 3).
4. Oxides of chromium (VI), bismuth (V), lead (IV), manganese (IV).
5. Hypochlorites (NaClO), chlorates (NaClO 3) and perchlorates (NaClO 4); nitrates (KNO 3).
6. Peroxides, superoxides, ozonides, organic peroxides, peroxoacids, all other substances containing the -O-O- group (for example, hydrogen peroxide - H 2 O 2, sodium peroxide - Na 2 O 2, potassium superoxide - KO 2).
7. Metal ions located on the right side of the voltage series: Au 3+, Ag +.

Typical reducing agents

1. Simple substances - metals: alkali and alkaline earth, Mg, Al, Zn, Sn.
2. Simple substances - non-metals: H 2, C.
3. Metal hydrides: LiH, CaH 2, lithium aluminum hydride (LiAlH 4), sodium borohydride (NaBH 4).
4. Hydrides of some non-metals: HI, HBr, H 2 S, H 2 Se, H 2 Te, PH 3, silanes and boranes.
5. Iodides, bromides, sulfides, selenides, phosphides, nitrides, carbides, nitrites, hypophosphites, sulfites.
6. Carbon monoxide (CO).

I would like to emphasize a few points:

1. I did not set myself the goal of listing all oxidizing and reducing agents. This is impossible, and it is not necessary.
2. The same substance can act as an oxidizing agent in one process, and as an oxidizing agent in another.
3. No one can guarantee that you will definitely encounter one of these substances in the C1 exam problem, but the probability of this is very high.
4. What is important is not mechanical memorization of formulas, but UNDERSTANDING. Try to test yourself: write out the substances from the two lists mixed together, and then try to independently separate them into typical oxidizing and reducing agents. Use the same considerations we discussed at the beginning of this article.

And now a small one test. I will offer you several incomplete equations, and you will try to find the oxidizing agent and the reducing agent. It is not necessary to add the right-hand sides of the equations yet.

Example 12. Determine the oxidizing agent and reducing agent in the ORR:

HNO3 + Zn = ...

CrO 3 + C 3 H 6 + H 2 SO 4 = ...

Na 2 SO 3 + Na 2 Cr 2 O 7 + H 2 SO 4 = ...

O 3 + Fe(OH) 2 + H 2 O = ...

CaH 2 + F 2 = ...

KMnO 4 + KNO 2 + KOH = ...

H 2 O 2 + K 2 S + KOH = ...

I think you completed this task without difficulty. If you have problems, read the beginning of this article again, work on the list of typical oxidizing agents.

“This is all wonderful!” the impatient reader will exclaim. “But where are the promised problems C1 with incomplete equations? Yes, in example 12 we were able to determine the oxidizing agent and the oxidizing agent, but that’s not the main thing. The main thing is to be able to COMPLETE the reaction equation, and can a list of oxidizing agents help us with this?"

Yes, it can, if you understand WHAT HAPPENS to typical oxidizing agents under different conditions. This is exactly what we will do now.

Sixth step: transformations of some oxidizing agents in different environments. "Fate" of permanganates, chromates, nitric and sulfuric acids

So, we must not only be able to recognize typical oxidizing agents, but also understand what these substances are converted into during the redox reaction. Obviously, without this understanding we will not be able to correctly solve problem 30. The situation is complicated by the fact that the products of interaction cannot be indicated UNIQUELY. It makes no sense to ask: “What will potassium permanganate turn into during the reduction process?” It all depends on many reasons. In the case of KMnO 4, the main one is the acidity (pH) of the medium. In principle, the nature of the recovery products may depend on:

1. reducing agent used during the process,
2. acidity of the environment,
3. concentrations of reaction participants,
4. process temperature.

We will not talk now about the influence of concentration and temperature (although inquisitive young chemists may recall that, for example, chlorine and bromine interact differently with an aqueous solution of alkali in the cold and when heated). Let's focus on the pH of the medium and the strength of the reducing agent.

The information below is simply something to remember. There is no need to try to analyze the causes, just REMEMBER the reaction products. I assure you, this may be useful to you at the Unified State Exam in Chemistry.

Products of the reduction of potassium permanganate (KMnO 4) in various media

Example 13. Complete the equations of redox reactions:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = ...
KMnO 4 + H 2 O + K 2 SO 3 = ...
KMnO 4 + KOH + K 2 SO 3 = ...

Solution. Guided by the list of typical oxidizing and reducing agents, we come to the conclusion that the oxidizing agent in all these reactions is potassium permanganate, and the reducing agent is potassium sulfite.

H 2 SO 4 , H 2 O and KOH determine the nature of the solution. In the first case, the reaction occurs in an acidic environment, in the second - in a neutral environment, in the third - in an alkaline environment.

Conclusion: in the first case, the permanganate will be reduced to Mn(II) salt, in the second - to manganese dioxide, in the third - to potassium manganate. Let's add the reaction equations:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + ...
KMnO 4 + KOH + K 2 SO 3 = K 2 MnO 4 + ...

What will potassium sulfite turn into? Well, naturally, into sulfate. It is obvious that the K in the composition of K 2 SO 3 simply has nowhere to oxidize further, the oxidation of oxygen is extremely unlikely (although, in principle, possible), but S(+4) easily turns into S(+6). The oxidation product is K 2 SO 4, you can add this formula to the equations:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + K 2 SO 4 + ...
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + K 2 SO 4 + ...
KMnO 4 + KOH + K 2 SO 3 = K 2 MnO 4 + K 2 SO 4 + ...

Our equations are almost ready. All that remains is to add substances that are not directly involved in OVR and set the coefficients. By the way, if you start from the second point, it may be even easier. Let's build, for example, an electronic balance for the last reaction

We put coefficient 2 in front of the formulas KMnO 4 and K 2 MnO 4; before the formulas of sulfite and potassium sulfate we mean coefficient. 1:

2KMnO 4 + KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + ...

On the right we see 6 potassium atoms, on the left - so far only 5. We need to correct the situation; put the coefficient 2 in front of the KOH formula:

2KMnO 4 + 2KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + ...

The final touch: on the left side we see hydrogen atoms, on the right there are none. Obviously, we urgently need to find some substance that contains hydrogen in the oxidation state +1. Let's get some water!

2KMnO 4 + 2KOH + K 2 SO 3 = 2K 2 MnO 4 + K 2 SO 4 + H 2 O

Let's check the equation again. Yes, everything is great!

“An interesting movie!” the vigilant young chemist will note. “Why did you add water at the last step? What if I want to add hydrogen peroxide or just H2 or potassium hydride or H2S? You added water because it Did you HAVE to add it or did you just feel like it?”

Well, let's figure it out. Well, firstly, we naturally do not have the right to add substances to the reaction equation at will. The reaction goes exactly the way it goes; as nature ordered. Our likes and dislikes cannot influence the course of the process. We can try to change the reaction conditions (increase the temperature, add a catalyst, change the pressure), but if the reaction conditions are given, its result can no longer depend on our will. Thus, the formula of water in the equation of the last reaction is not my desire, but a fact.

Secondly, you can try to equalize the reaction in cases where the substances you listed are present instead of water. I assure you: in no case will you be able to do this.

Thirdly, options with H 2 O 2, H 2, KH or H 2 S are simply unacceptable in this case for one reason or another. For example, in the first case, the oxidation state of oxygen changes, in the second and third - of hydrogen, and we agreed that the oxidation state will change only for Mn and S. In the fourth case, sulfur generally acted as an oxidizing agent, and we agreed that S - reducing agent. In addition, potassium hydride is unlikely to “survive” in an aqueous environment (and the reaction, let me remind you, takes place in an aqueous solution), and H 2 S (even if this substance were formed) will inevitably enter into a solution with KOH. As you can see, knowledge of chemistry allows us to reject these substances.

"But why water?" - you ask.

Yes, because, for example, in this process (as in many others) water acts as a solvent. Because, for example, if you analyze all the reactions you wrote in 4 years of studying chemistry, you will find that H 2 O occurs in almost half of the equations. Water is generally a fairly “popular” compound in chemistry.

Please understand that I am not saying that every time in problem 30 you need to “send hydrogen somewhere” or “take oxygen from somewhere,” you need to grab water. But this will probably be the first substance you should think about.

Similar logic is used for reaction equations in acidic and neutral media. In the first case, you need to add the formula of water to the right side, in the second - potassium hydroxide:

KMnO 4 + H 2 SO 4 + K 2 SO 3 = MnSO 4 + K 2 SO 4 + H 2 O,
KMnO 4 + H 2 O + K 2 SO 3 = MnO 2 + K 2 SO 4 + KOH.

The arrangement of coefficients should not cause the slightest difficulty for experienced young chemists. Final answer:

2KMnO 4 + 3H 2 SO 4 + 5K 2 SO 3 = 2MnSO 4 + 6K 2 SO 4 + 3H 2 O,
2KMnO 4 + H 2 O + 3K 2 SO 3 = 2MnO 2 + 3K 2 SO 4 + 2KOH.

In the next part we will talk about the reduction products of chromates and dichromates, nitric and sulfuric acids.

How to solve problems C1 (36) on the Unified State Exam in chemistry. Part I

Problem No. 36 on the Unified State Exam in Chemistry is devoted to the topic “Oxidation-reduction reactions.” Previously, this type of task was included in Unified State Exam option under number C1.

The meaning of task C1: it is necessary to arrange the coefficients in the reaction equation using the electronic balance method. Usually, only the left side of the equation is given in the problem statement; the student must independently complete the right side.

A complete solution to the problem is worth 3 points. One point is given for determining the oxidizing agent and reducing agent, another one is given directly for constructing the electronic balance, the last one is for the correct arrangement of coefficients in the reaction equation.

In my opinion, the most difficult thing in this process is the first step. Not everyone is able to correctly predict the outcome of a reaction. If the interaction products are indicated correctly, all subsequent stages are a matter of technology.

First step: remember the oxidation states

We must start with the concept element oxidation state. If you are not yet familiar with this term, refer to the Oxidation State section in your chemistry reference book. You must learn to confidently determine the oxidation states of all elements in inorganic compounds and even in the simplest organic substances. Without a 100% understanding of this topic, moving forward is pointless.

Step two: oxidizing agents and reducing agents. Redox reactions

I want to remind you that everything chemical reactions in nature can be divided into two types: redox and occurring without changing oxidation states.

During the redox reaction (this is the abbreviation we will use further for redox reactions), some elements change their oxidation states.

Example 1. Consider the reaction of sulfur with fluorine:

S + 3F 2 = SF 6.

Arrange the oxidation states of all elements yourself. We see that the oxidation state of sulfur increases (from 0 to +6), and the oxidation state of fluorine decreases (from 0 to -1). Conclusion: S is a reducing agent, F 2 is an oxidizing agent. During the process, sulfur is oxidized and fluorine is reduced.

Example 2. Let's discuss the reaction of manganese (IV) oxide with hydrochloric acid:

MnO 2 + 4HCl = MnCl 2 + Cl 2 + 2H 2 O.

During the reaction, the oxidation state of manganese decreases (from +4 to +2), and the oxidation state of chlorine increases (from -1 to 0). Conclusion: manganese (in the composition of MnO 2) is an oxidizing agent, chlorine (in the composition of HCl is a reducing agent). Chlorine is oxidized, manganese is reduced.

Please note that in the last example, not all chlorine atoms changed oxidation state. This did not influence our conclusions in any way.

Example 3. Thermal decomposition of ammonium bichromate:

(NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O.

We see that both the oxidizing agent and the reducing agent are part of one “molecule”: chromium changes its oxidation state from +6 to +3 (i.e., it is an oxidizing agent), and nitrogen - from -3 to 0 (hence, nitrogen is reducing agent).

Example 4. Interaction of nitrogen dioxide with aqueous alkali solution:

2NO 2 + 2NaOH = NaNO 3 + NaNO 2 + H 2 O.

Having arranged the oxidation states (I hope you do this without difficulty!), we discover a strange picture: the oxidation state of only one element changes - nitrogen. Some N atoms increase their oxidation state (from +4 to +5), while others decrease them (from +4 to +3). In fact, there is nothing strange about this! In this process, N(+4) is both an oxidizing agent and a reducing agent.

Let's talk a little about the classification of redox reactions. Let me remind you that all OVRs are divided into three types:

• 1) intermolecular ORRs (the oxidizing agent and the reducing agent are contained in different molecules);
• 2) intramolecular ORRs (the oxidizing agent and the reducing agent are in one molecule);
• 3) disproportionation reactions (an oxidizing agent and a reducing agent are atoms of the same element with the same initial oxidation state in the composition of one molecule).

I think that, based on these definitions, you can easily understand that the reactions from examples 1 and 2 relate to intermolecular ORR, the decomposition of ammonium dichromate is an example of intramolecular ORR, and the interaction of NO 2 with alkali is an example of a disproportionation reaction.

Step three: we begin to master the electronic balance method

To check how well you have mastered the previous material, I’ll ask you a simple question: “Can you give an example of a reaction in which oxidation occurs but there is no reduction, or, conversely, there is oxidation but no reduction?”

Correct answer: "No, you can't!"

Indeed, let the oxidation state of element X increase during the reaction. This means that X donates electrons. But to whom? After all, electrons cannot simply evaporate, disappear without a trace! There is some other element Y whose atoms will accept these electrons. Electrons have a negative charge, therefore the oxidation state of Y will decrease.

Conclusion: if there is a reducing agent X, then there will certainly be an oxidizing agent Y! Moreover, the number of electrons given up by one element will be exactly equal to the number of electrons accepted by another element.

It is on this fact that it is based electronic balance method, used in task C1.

Let's begin to master this method with examples.

Example 4

C + HNO 3 = CO 2 + NO 2 + H 2 O

electronic balance method.

Solution. Let's start by determining oxidation states (do it yourself!). We see that during the process two elements change their oxidation states: C (from 0 to +4) and N (from +5 to +4).

It is obvious that carbon is a reducing agent (oxidized), and nitrogen (+5) (in the composition nitric acid) is an oxidizing agent (reduced). By the way, if you correctly identified the oxidizing agent and the in-tel, you are already guaranteed 1 point for problem N 36!

Now the fun begins. Let's write the so-called half-reactions of oxidation and reduction:

The carbon atom gives up 4 electrons, the nitrogen atom gains 1 electron. The number of electrons given is not equal to the number of electrons received. This is bad! The situation needs to be corrected.

Let's "multiply" the first half-reaction by 1, and the second by 4.

Now everything is fine: for one carbon atom (giving 4 e) there are 4 nitrogen atoms (each of which takes one e). The number of electrons given is equal to the number of electrons received!

What we have just written is actually called electronic balance. If you write this balance correctly on a real Unified State Exam in chemistry, you are guaranteed 1 more point for problem C1.

The last stage: it remains to transfer the obtained coefficients into the reaction equation. Before the formulas C and CO 2 we do not change anything (since the coefficient 1 is not put in the equation), before the formulas HNO 3 and NO 2 we put a four (since the number of nitrogen atoms on the left and right sides of the equation should be equal to 4) :

C + 4HNO 3 = CO 2 + 4NO 2 + H 2 O.

It remains to do one last check: we see that the number of nitrogen atoms is the same on the left and right, the same applies to the C atoms, but there are still problems with hydrogen and oxygen. But everything is easy to fix: we put a factor of 2 in front of the formula H 2 O and get the final answer:

C + 4HNO 3 = CO 2 + 4NO 2 + 2H 2 O.

That's all! The problem is solved, the coefficients are set, and we get one more point for the correct equation. Result: 3 points for a perfectly solved problem C 1. Congratulations on that!

Example 5. Arrange the coefficients in the reaction equation

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + I 2 + H 2 O

electronic balance method.

Solution. Arrange the oxidation states of all elements yourself. We see that during the process two elements change their oxidation states: S (from +6 to -2) and I (from -1 to 0).

Sulfur (+6) (in sulfuric acid) is an oxidizing agent, and iodine (-1) in NaI is a reducing agent. During the reaction, I(-1) is oxidized, S(+6) is reduced.

We write down the half-reactions of oxidation and reduction:

pay attention to important point: There are two atoms in an iodine molecule. “Half” of the molecule cannot participate in the reaction, so in the corresponding equation we write not I, but precisely I 2.

Let us “multiply” the first half-reaction by 4, and the second by 1.

The balance is built, for every 8 electrons given there are 8 electrons received.

We transfer the coefficients into the reaction equation. Before the formula I 2 we put 4, before the formula H 2 S we mean the coefficient 1 - this, I think, is obvious.

NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

But further questions may arise. Firstly, it would be incorrect to put a four in front of the NaI formula. Indeed, already in the oxidation half-reaction itself, the symbol I is preceded by a coefficient of 2. Consequently, not 4, but 8 should be written on the left side of the equation!

8NaI + H 2 SO 4 = Na 2 SO 4 + H 2 S + 4I 2 + H 2 O

Secondly, in such a situation, graduates often put a factor of 1 in front of the formula of sulfuric acid. They reason like this: “In the reduction half-reaction, a coefficient of 1 was found, this coefficient refers to S, which means that the formula of sulfuric acid must be preceded by a unit.”

These reasonings are wrong! Not all sulfur atoms changed their oxidation state; some of them (in the composition of Na 2 SO 4) retained the oxidation state +6. These atoms are not taken into account in the electronic balance and coefficient 1 has nothing to do with them.

All this, however, will not prevent us from bringing the decision to completion. It is only important to understand that in further discussions we no longer rely on the electronic balance, but simply on common sense. So, I remind you that the coefficients for H 2 S, NaI and I 2 are “frozen” and cannot be changed. But the rest - it is possible and necessary.

On the left side of the equation there are 8 sodium atoms (in NaI), on the right there are only 2 atoms so far. We put a factor of 4 in front of the sodium sulfate formula:

8NaI + H 2 SO 4 = 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

Only now can you equalize the number of S atoms. There are 5 of them on the right, therefore, you need to put a factor of 5 in front of the formula of sulfuric acid:

8NaI + 5H 2 SO 4 = 4Na 2 SO 4 + H 2 S + 4I 2 + H 2 O.

The last problem: hydrogen and oxygen. Well, I think you yourself guessed that the coefficient 4 is missing in front of the water formula on the right side:

8NaI + 5H 2 SO 4 = 4Na 2 SO 4 + H 2 S + 4I 2 + 4H 2 O.

We check everything carefully again. Yes everything is correct! The problem is solved, we received our rightful 3 points.

So, in examples 4 and 5 we discussed in detail algorithm for solving problem C1. Your solution to a real exam problem must include the following points:

• 1) oxidation states of ALL elements;
• 2) indication of the oxidizing agent and reducing agent;
• 3) electronic balance scheme;
• 4) the final reaction equation with coefficients.

A few comments about the algorithm.

1. The oxidation states of all elements on the left and right sides of the equation must be indicated. Everyone, not just the oxidizing agent and the reducing agent!

2. The oxidizing agent and the reducing agent must be clearly and clearly indicated: element X (+...) in the composition... is an oxidizing agent and is reduced; element Y(...) in the composition... is a reducing agent and is oxidized. Not everyone will be able to decipher the inscription in small handwriting “ok. in” under the formula of sulfuric acid as “sulfur (+6) in the composition of sulfuric acid is an oxidizing agent, reduced.”

Don't skimp on letters! You don’t put an ad in the newspaper: “Led room with all amenities.”

3. The electronic balance diagram is just a diagram: two half-reactions and the corresponding coefficients.

4. No one needs detailed explanations of exactly how you placed the coefficients in the equation on the Unified State Exam. It is only necessary that all the numbers are correct, and the entry itself is made in legible handwriting. Be sure to check yourself several times!

And once again regarding the assessment of task C1 on the Unified State Exam in chemistry:

• 1) determination of the oxidizing agent (oxidizing agents) and reducing agent (reducing agents) - 1 point;
• 2) electronic balance scheme with correct coefficients - 1 point;
• 3) the basic reaction equation with all coefficients - 1 point.

Result: 3 points for complete solution of problem No. 36.

I'm sure you understand what the idea behind the electronic balance method is. We understood in basic terms how the solution to example C1 is constructed. In principle, everything is not so difficult!

Unfortunately, on a real Unified State Exam in chemistry the following problem arises: the reaction equation itself is not given in full. That is, the left side of the equation is present, but on the right there is either nothing at all or the formula of one substance is indicated. You will have to complete the equation yourself, based on your knowledge, and only then begin arranging the coefficients.

This can be quite difficult. There are no universal recipes for writing equations. In the next part we will discuss this issue in more detail and look at more complex examples.

Copyright Repetitor2000.ru, 2000-2015

Redox reactions

For completing the task correctly you will receive 2 points. It takes approximately 10-15 minutes.

To complete task 30 in chemistry you must:

• know what it is
• be able to write equations of redox reactions